Does (1/n+1/ni) converge to zero?

1. The problem statement, all variables and given/known data
Does the sequence [itex]\frac{1}{n}[/itex]+[itex]\frac{1}{n}[/itex]i converge to zero with respect to the metric d(z,w) = |z|+|w|.

3. The attempt at a solution
So I realise I want √([itex]\frac{1}{n}[/itex]2+[itex]\frac{1}{n}[/itex]2)=[itex]\frac{√2}{n}[/itex] to be less than ε. So for all n greater than [itex]\frac{√2}{ε}[/itex] the order [itex]\frac{√2}{n}[/itex]<ε will hold.

So using the definition for all ε>0 there exists and nε=[itex]\frac{√2}{ε}[/itex] such that m>n implies [itex]\frac{√2}{n}[/itex]<ε.

Is my technique correct? Is finding a function for n in terms of ε that always maintains the equality sufficient for a proof? If this is the case and ε is a real number how to we ensure that n is a natural number? Does n have to be natural number or can that condition just be put on m.

1. The problem statement, all variables and given/known data
Does the sequence [itex]\frac{1}{n}[/itex]+[itex]\frac{1}{n}[/itex]i converge to zero with respect to the metric d(z,w) = |z|+|w|.

3. The attempt at a solution
So I realise I want √([itex]\frac{1}{n}[/itex]2+[itex]\frac{1}{n}[/itex]2)=[itex]\frac{√2}{n}[/itex] to be less than ε. So for all n greater than [itex]\frac{√2}{ε}[/itex] the order [itex]\frac{√2}{n}[/itex]<ε will hold.

So using the definition for all ε>0 there exists and nε=[itex]\frac{√2}{ε}[/itex] such that m>n implies [itex]\frac{√2}{n}[/itex]<ε.

Is my technique correct? Is finding a function for n in terms of ε that always maintains the equality sufficient for a proof? If this is the case and ε is a real number how to we ensure that n is a natural number? Does n have to be natural number or can that condition just be put on m.

Looks ok. You generally choose n by just saying that it's an integer greater than [itex]\sqrt{\frac{2}{ε}}[/itex]. No need to worry about whether [itex]\sqrt{\frac{2}{ε}}[/itex] itself is an integer.